The freezing point of water is 0.00 °C at 1 atmosphere.

A student dissolves 14.38 grams of chromium(II) iodide ,  (305.8 g/mol), in 264.4 grams of water

This is a freezing point depression problem where we can use the formula:

\[
\Delta T_f = i \cdot K_f \cdot m
\]

Where:
- \(\Delta T_f\) is the freezing point depression,
- \(i\) is the van't Hoff factor (number of particles the solute dissociates into),
- \(K_f\) is the freezing point depression constant for water (1.86 °C·kg/mol),
- \(m\) is the molality of the solution (moles of solute per kilogram of solvent).

### Step 1: Calculate the number of moles of chromium(II) iodide.
The molar mass of chromium(II) iodide \((CrI_2)\) is given as 305.8 g/mol, and the student dissolved 14.38 grams.

\[
\text{moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} = \frac{14.38}{305.8} \approx 0.0470 \text{ mol}
\]

### Step 2: Calculate the molality \(m\) of the solution.
Molality is defined as moles of solute per kilogram of solvent. The mass of the solvent (water) is 264.4 grams, which is equivalent to 0.2644 kg.

\[
m = \frac{0.0470 \text{ mol}}{0.2644 \text{ kg}} \approx 0.1778 \text{ mol/kg}
\]

### Step 3: Determine the van't Hoff factor \(i\).
Chromium(II) iodide \((CrI_2)\) dissociates into 1 \(Cr^{2+}\) ion and 2 \(I^-\) ions, so \(i = 3\).

### Step 4: Calculate the freezing point depression \(\Delta T_f\).

\[
\Delta T_f = i \cdot K_f \cdot m = 3 \cdot 1.86 \cdot 0.1778 \approx 0.993 \text{ °C}
\]

### Step 5: Determine the new freezing point.
The freezing point of pure water is 0.00°C. Subtract the freezing point depression from this:

\[
\text{New freezing point} = 0.00 \, ^\circ\text{C} - 0.993 \, ^\circ\text{C} \approx -0.993 \, ^\circ\text{C}
\]

### Conclusion:
The new freezing point of the solution is approximately \(-0.993 \, ^\circ\text{C}\).

Would you like further details or clarifications on any step?

Here are 5 related questions to expand on this:
1. How does the van't Hoff factor \(i\) change for different compounds?
2. What is the significance of molality in colligative properties?
3. How do you calculate boiling point elevation using similar principles?
4. How would a non-electrolyte solute affect the freezing point differently?
5. What role does the solvent's identity play in freezing point depression?

**Tip:** Colligative properties (like freezing point depression) depend on the number of particles in the solution, not the identity of the solute.

The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 14.38 grams of chromium(II) iodide (305.8 g/mol) in 264.4 grams of water. What is the new freezing point of the solution?

Learn how to calculate the freezing point depression when 14.38 grams of chromium(II) iodide is dissolved in 264.4 grams of water. This problem covers colligative properties and includes detailed steps on calculating molality and applying the freezing point depression formula. Suitable for high school students in Grades 10-12.

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